the weight of oranges growing in an orchard is normally distributed with a mean weight of 5.5 oz. and a…

the weight of oranges growing in an orchard is normally distributed with a mean weight of 5.5 oz. and a standard deviation of 0.5 oz. using the empirical rule, what percentage of the oranges from the orchard weigh between 5 oz. and 6 oz.?

the weight of oranges growing in an orchard is normally distributed with a mean weight of 5.5 oz. and a standard deviation of 0.5 oz. using the empirical rule, what percentage of the oranges from the orchard weigh between 5 oz. and 6 oz.?

Answer

Explanation:

Step1: Recall the empirical rule

The empirical rule for a normal - distribution states that about 68% of the data lies within 1 standard deviation of the mean, about 95% lies within 2 standard deviations of the mean, and about 99.7% lies within 3 standard deviations of the mean. The mean $\mu = 5.5$ oz and the standard deviation $\sigma=0.5$ oz.

Step2: Calculate the number of standard deviations

For the lower value $x_1 = 5$ oz, the z - score $z_1=\frac{x_1-\mu}{\sigma}=\frac{5 - 5.5}{0.5}=\frac{- 0.5}{0.5}=-1$. For the upper value $x_2 = 6$ oz, the z - score $z_2=\frac{x_2-\mu}{\sigma}=\frac{6 - 5.5}{0.5}=\frac{0.5}{0.5}=1$.

Step3: Apply the empirical rule

The values 5 oz and 6 oz are 1 standard deviation below and 1 standard deviation above the mean respectively. According to the empirical rule, the percentage of data within $z=-1$ and $z = 1$ is 68%.

Answer:

68%